Как решать:

5sin2x=0

-4cos (p/4-0,1x) = 0

5sin (2x) = 0, sin (2x) = 0, 2x = arc sin 0 = πk, k ∈ Z.

x = (πk/2), k ∈ Z.

-4cos ((π / 4) - 0,1x) = 0, cos ((π/4) - 0,1x) = 0, (π/4) - 0,1x = arc cos 0.

(π/4) - 0,1x = (π/2) + πk, k ∈ Z.

x = ((π/4) - (π/2) - πk) * 10 = (-10π/4) - 10πk = (-5π/2) - 10π, k ∈ Z.